To find $f'''(x)$, the third derivative of the function $f(x) = -\frac{5}{6}x^3 + x^2 + \frac{3}{7} - \frac{1}{3}x^5$, we need to differentiate $f(x)$ three times. Let’s proceed with each derivative step-by-step.

1. First Derivative $f'(x)$: $f(x) = -\frac{5}{6}x^3 + x^2 + \frac{3}{7} - \frac{1}{3}x^5$ Differentiate each term:

   • The derivative of $-\frac{5}{6}x^3$ is $-\frac{5}{6} \cdot 3x^2 = -\frac{5}{2}x^2$.
   • The derivative of $x^2$ is $2x$.
   • The derivative of $\frac{3}{7}$ (a constant) is $0$.
   • The derivative of $-\frac{1}{3}x^5$ is $-\frac{1}{3} \cdot 5x^4 = -\frac{5}{3}x^4$.

   So, $f'(x) = -\frac{5}{2}x^2 + 2x - \frac{5}{3}x^4$

2. Second Derivative $f''(x)$: Differentiate $f'(x)$:

   • The derivative of $-\frac{5}{2}x^2$ is $-\frac{5}{2} \cdot 2x = -5x$.
   • The derivative of $2x$ is $2$.
   • The derivative of $-\frac{5}{3}x^4$ is $-\frac{5}{3} \cdot 4x^3 = -\frac{20}{3}x^3$.

   So, $f''(x) = -5x + 2 - \frac{20}{3}x^3$

3. Third Derivative $f'''(x)$: Differentiate $f''(x)$:

   • The derivative of $-5x$ is $-5$.
   • The derivative of $2$ (a constant) is $0$.
   • The derivative of $-\frac{20}{3}x^3$ is $-\frac{20}{3} \cdot 3x^2 = -20x^2$.

   So, $f'''(x) = -5 - 20x^2$

Final Answer: $f'''(x) = -20x^2 - 5$

Would you like further details on each step, or do you have any questions?

Here are some related questions to deepen your understanding:

1. What is the fourth derivative of $f(x)$ based on this answer?
2. How would $f'''(x)$ change if $f(x)$ included higher-degree terms, like $x^6$?
3. Can you explain why constants disappear when taking derivatives?
4. How does the power rule apply to each term in $f(x)$ when differentiating?
5. What would happen to $f'(x)$ if you only differentiated the constant term $\frac{3}{7}$?

Tip: Always remember that each differentiation reduces the degree of each term by one.